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2 years ago

A child jumps on a trampoline, which of the following causes the child to rise in the air?

Physics

1 answer:

weeeeeb [17]2 years ago

4 0

Answer:

Force pull the child up. With the springs and the weight when you jump.

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What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?

EleoNora [17]

Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)

4 0

2 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

5 0

3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector $X = 7.8*Cos(50) = 5.01 Km$ y $y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km)$ so that $Tan (\alpha ) = \frac{0.375}{5.01} = 4.28$ ; $\alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree)$ to get how far we use Pythagorean theorem so $R^{2} = x^{2}+y^{2}$ so that $R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)$